Descartes’ Rule of Signs

Descartes’ rule of signs predicts how many positive and negative real zeros a polynomial can have. Count the sign changes in f(x), written in descending order: the number of positive real zeros equals that count or is less by an even number. Count the sign changes in f(-x) the same way to get the possible number of negative real zeros.

Before this lesson: You should know the Fundamental Theorem of Algebra, the Rational Root Theorem and Even and Odd Functions.

What the Rule Tells You

Before you hunt for zeros, it helps to know how many to expect. The fundamental theorem of algebra says a polynomial of degree n has n zeros, counting multiplicity. Some are positive, some negative, some may be 0, and the rest are nonreal.

Descartes’ rule splits up the real ones. A sign change happens when two neighboring coefficients have opposite signs. Each positive zero forces at least one sign change, so the number of sign changes is the most positive zeros there can be.

Nonreal zeros come in conjugate pairs, like 2 + i and 2 - i. That is why the count drops by 2 at a time, never by 1.

Descartes’ Rule of Signs

1. Write f(x) in descending powers. Skip any missing terms; they don’t count.
2. Count the sign changes from one coefficient to the next. The number of positive real zeros is that number, or less by 2, 4, and so on.
3. Find f(-x): change the sign of every term with an odd power of x.
4. Count the sign changes in f(-x). The number of negative real zeros is that number, or less by an even number.
5. If the constant term is 0, factor out the power of x first. Zero is neither positive nor negative.

Zeros are counted with multiplicity, so a double zero counts as two.

Worked Examples

Example 1: Count both kinds

Use Descartes’ rule on f(x) = x^3 - 4x^2 + x + 6.

+, -, +, +

Count the changes: + to - (1), - to + (2), + to + (none). Two sign changes, so 2 or 0 positive zeros

(-x)^3 - 4(-x)^2 + (-x) + 6 = -x^3 - 4x^2 - x + 6

List its signs: -, -, -, +. One sign change, so exactly 1 negative zero

Check by factoring: f(x) = (x + 1)(x - 2)(x - 3), with zeros -1, 2, 3

Answer: 2 or 0 positive zeros and exactly 1 negative zero. The actual zeros, 2 and 3 positive and −1 negative, fit.

Example 2: No real zeros

Use Descartes’ rule on f(x) = 2x^4 + 3x^2 + 5.

+, +, +

f(-x) = 2x^4 + 3x^2 + 5

The constant is 5, not 0, so 0 is not a zero.

Answer: no real zeros. All 4 zeros are nonreal.

Example 3: A table of possibilities

List the possible numbers of positive, negative and nonreal zeros of f(x) = x^5 - 3x^3 + 2x - 7.

Signs of f(x): +, -, +, -. Three changes, so 3 or 1 positive zeros

-x^5 + 3x^3 - 2x - 7

Signs of f(-x): -, +, -, -. Two changes, so 2 or 0 negative zeros

The degree is 5, so the nonreal count makes the total 5: with 3 positive and 2 negative, 5 - 3 - 2 = 0 nonreal

Fill in the other rows the same way

Answer: four possible combinations, shown in the table. Every row has an even number of nonreal zeros.

Example 4: A zero at the origin

Use Descartes’ rule on f(x) = x^4 - 5x^2 + 4x.

f(x) = x(x^3 - 5x + 4)

+, -, +

g(-x) = -x^3 + 5x + 4

-, +, +

g(1) = 1 - 5 + 4 = 0

g(x) = (x - 1)(x^2 + x - 4)

x = \frac{-1 \pm \sqrt{17}}{2}

Answer: one zero at 0, 2 positive zeros (1 and about 1.56) and 1 negative zero (about −2.56), just as the rule allowed.

Example 5: Word problem with a box

Squares of side x inches are cut from the corners of a 12-by-10 inch sheet, and the sides are folded up to make an open box. The volume is V(x) = x(12 - 2x)(10 - 2x). How many cut sizes could give a volume of 96 cubic inches?

V(x) = 4x^3 - 44x^2 + 120x

4x^3 - 44x^2 + 120x - 96 = 0

x^3 - 11x^2 + 30x - 24 = 0

Signs: +, -, +, -. Three changes, so 3 or 1 positive solutions

Test x = 2: 8 - 44 + 60 - 24 = 0, so 2 is a solution. Dividing gives (x - 2)(x^2 - 9x + 12) = 0

Solve x^2 - 9x + 12 = 0: x = \frac{9 \pm \sqrt{33}}{2}, about 7.37 and 1.63

A cut must be less than 5 inches, since the sheet is only 10 inches wide, so throw out 7.37

Answer: The rule allowed 3 positive solutions, and there are 3, but only 2 make sense: x = 2 inches or about 1.63 inches.

Common Mistakes

Related Methods

Rational root theorem

The rule tells you how many zeros to look for and on which side of 0. The rational root theorem lists the candidates, and synthetic division tests them. If there is only 1 negative zero, you can stop looking once you find it.

Check with a graph

A graph shows the real zeros as x-intercepts, like the picture above. Use it to decide which row of the possibilities table is true.

Nonreal zeros

Whatever is left over after the real zeros comes in conjugate pairs, the same way complex solutions of quadratic equations do.

Practice Problems

1. Use Descartes’ rule on f(x) = x^3 + 2x^2 - 5x - 6.
2. Use Descartes’ rule on f(x) = 3x^4 - 2x^3 + x^2 - 5x + 1.
3. Use Descartes’ rule on f(x) = x^4 + x^2 + 1.
4. Use Descartes’ rule on f(x) = -x^3 + 4x - 1.
5. Use Descartes’ rule on f(x) = x^5 + x^3 + x.
6. How many positive, negative and nonreal zeros does f(x) = x^3 - 1 have?
7. List every possible combination of positive, negative and nonreal zeros for f(x) = x^4 - 2x^3 - x^2 + 2x + 5.
8. Explain why f(x) = x^6 + 3x^2 + 2 has no real zeros.
9. Use Descartes’ rule on f(x) = 2x^3 - 5x^2 + 4x - 1, then check with the factored form (x - 1)^2(2x - 1).
10. A company’s profit is P(x) = -x^3 + 10x^2 - 21x - 18 thousand dollars when it sells x thousand units. How many break-even points (zeros) with x \gt 0 are possible? Check with P(6).

FAQ

How do you use Descartes’ rule of signs?

Count the sign changes in f(x) for the possible positive real zeros. Then count the sign changes in f(-x) for the possible negative real zeros. Each count can drop by any even number.

Why does the number of zeros go down by 2?

Nonreal zeros of a polynomial with real coefficients come in conjugate pairs. Any zeros missing from the count are nonreal, and they come two at a time.

Do zeros with multiplicity count more than once?

Yes. A double zero counts as two zeros in Descartes’ rule, the same way it does in the fundamental theorem of algebra.

Does Descartes’ rule give the exact number of zeros?

Only when the count is 0 or 1. Otherwise, it gives a list of possibilities, and you need more work, like a graph or synthetic division, to decide.

Summary

• Sign changes in f(x) give the most positive real zeros possible.
• Sign changes in f(-x) give the most negative real zeros possible.
• Each count may drop by 2, 4 and so on, because nonreal zeros come in pairs.
• Skip missing terms, and factor out x if the constant is 0.
• Use it with the rational root theorem and a graph to find the zeros.

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